Suppose then taking norms (remember, ) we find that . W.l.o.g. we have either or and .

Now, simply prove that no element has norm 2 and that implies that is a unit ( ).

It is probably good revision making up a few of these yourselves, especially where they map onto the integers not just the natural numbers. Also, I once did an exam that asked me lots of these types of questions - I would advise you to write something like " is a norm function because..." at the very start. I did that little proof at the start of every such question. It nearly killed me, and it meant didn't quite have time to finish the paper properly...

It is the sum of two positive things, if then just the right term alone is too big so equality is not possible. Thus b=0.

But this means .
Is there an integer such that when you square it you get two? Not a chance. About a million ways to show that take your pick is not rational so its not an integer.

Thus the only cases that are possible were case 1 and 2.
In both of these cases we showed either or is a unit.

What is the point of all this? Really you should show all 4 of the elements of the two factorizations I gave you are irreducible. Then you would have shown R is an integral domain that is in fact not a Unique Factorization Domain as you have just found two unique factorizations of 6 into irreducibles.